Consider the tabulated data showing the initial rate of a reaction (A → products) at several different concentrations of A. What is the order of the reaction? Write a rate law for the reaction including the value of the rate constant, k.

Video transcript

hey everyone in this example we're given a table below with the initial rate for the reaction of a producing be at different concentrations of A. And we need to determine the complete rate law for the reaction. So what we should recognize is that this first row represents rate one and the second row represents rate to and what we should recall is that we would take Rate two divided by rate one. To solve for our rate constant K. Which we would take by multiplying this by the initial concentration of our reactant A. To the end order. Which we have to solve for for rate two divided by the rate constant K. Times the initial concentration of are reacting A. To the end order for rate one. And so now we're going to utilize the values given in our data set. So what we should have is for rate to we have a value of 0.400 molar per second. This is divided by R. Value for rate one. Which has an initial rate of 0.108 polarities per second. This is set equal to our right hand side. Where in our numerator we plug in K. Times our concentration of a. At 0.0 400 Polarities per second. And according to the data set that's 0.250 More clarity as our concentration of a. To the order for the second rate. And then in our denominator we have the rate constant K. multiplied by the concentration of a react in a at the initial rate one which is 0.130 moller. And this is to the order for the initial rate one. So just so it's clear we can just actually understand that this is for the second rate and this is for the first rate. So we don't have to write those values in. Now. We just want to simplify to solve for N. So simplifying both sides. We're gonna get on the left hand side a value of 3.7037 are units of polarities per second, cancel out for both sides. And then this is going to be equal to the right hand side. Where we'll get a value of 1.9 to 31 to the 10th power. So two sulfur n. We would say that N. is equal to in our numerator, the natural log of 3.7037 divided by the natural log of 1.9231. And this would give us a value for N. Equal to two. And so we would say that therefore our reaction is second order. And now we want to recall our rate law for a second order reaction where we would say that our rate is equal to our rate constant K. Times the concentration of our react in A. Which is raised to the order being to now. And so now we can just plug in what we know from our dataset. So what we would get is that our rate initially for rate one is 0.108 polarities per second according to our data set. This is set equal to our rate constant K. Which we multiply by the initial concentration which corresponds to our data set at this initial rate being 0.130 moller. And this is still raised to the second power because we have a second order reaction. So we would simplify this to sell for the rate. And what we would get in the next line is 0.0108 polarities per second is equal to the rate constant K. multiplied by our concentration squared which is going to give us a value of 0.0169. And so now we're just going to further isolate for the rate constant K by dividing both sides by 0.0169. And these are in units of still moller here. So I should fill that in there and our units of polarity will now cancel out where we will get for our rate constant K. A value equal to 0.639. And just to go back to the units, this would be squared polarity meaning just to correct this step here, we would still have polarity squared being divided on both sides. However, now we're just going to be left with an exponent of one, meaning we just have one unit of polarity left. And so that means that our final units are going to be inverse polarity times inverse seconds. Since everything is in the numerator now. And so now we can therefore right out our rate where we would say that our rate Is equal to our rate constant, which we now know is equal to 0.639 inverse polarity times inverse seconds multiplied by our concentration of our reactant. A and this here is going to complete our answer as or complete this example as our final answer. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below. And just to be clear, this is for a second order reaction. So that's also a part of our answer. And I will see everyone in the next practice video.